\documentclass[aps,twocolumn]{revtex4}
\usepackage{graphicx}
\usepackage{amssymb,amsfonts,amsmath,amsthm}
\usepackage{chemarr}
\usepackage{bm}
\usepackage{pslatex}
\usepackage{mathptmx}
\usepackage{xfrac}

%% concentration notations
\newcommand{\mymat}[1]{\boldsymbol{#1}}
\newcommand{\mytrn}[1]{{#1}^{\mathsf{T}}}
\newcommand{\myvec}[1]{\overrightarrow{#1}}
\newcommand{\mygrad}{\vec{\nabla}}
\newcommand{\myhess}{\mathcal{H}}

\begin{document}

\section{Setup}
Let $\vec{F}\left(\vec{X}\right)$ be a system of $n$ equations with $n$ variables.
Starting from $\vec{X}$, what is the step to take ?
In any case, for a small step $\vec{h}$ we have
\begin{equation}
	\vec{F}\left(\vec{X}+\vec{h}\right) = \vec{F} + \mymat{J} \vec{h}
\end{equation}
We define the associated objective function
\begin{equation}
	f\left(\vec{X}\right) = \frac{1}{2} \vec{F}^2
\end{equation}
We notice that
\begin{equation}
	\mygrad f = \mytrn{\mymat{J}}\vec{F}
\end{equation}
and that
\begin{equation}
	\myhess_f = \mytrn{\mymat{J}}\mymat{J} + \partial_{\vec{X}}\mymat{J} \otimes \vec{F}.
\end{equation}


\section{Well-conditioned Jacobian}
In that case, we can compute the full Newton's step
\begin{equation}
	\vec{h} = - \mymat{J}^{-1} \vec{F}
\end{equation}
Let us now define
\begin{equation}
	\phi(\lambda) =  f \left(\vec{X} + \lambda \vec{h}\right)
\end{equation}
The initial descent rate of $\phi$ is
\begin{equation}
	\phi'(0) = \mygrad f \cdot \vec{h} \leq 0
\end{equation}
We must control the descent rate so that we don't jump too far or decrease too slowly.\\
In the fully parabolic case, the 0 of $\phi$ is found by following \emph{half} the descent rate.
Thus we need the control parameter
\begin{equation}
	0 < \alpha < \frac{1}{2},
\end{equation}
the descent rate
\begin{equation}
	\rho = - \mygrad f \cdot \vec{h} \geq 0,
\end{equation}
and accept a step only if
\begin{equation}
	\label{ok}
	\phi(\lambda) \leq \phi(0) - \alpha \lambda \rho.
\end{equation}
In the algorithm, we always try $\lambda=1$ and test the convergence
if the condition \eqref{ok} is fulfilled.
Otherwise, we must backtrack to an acceptable solution, that can't be tested for convergence (not in
the parabolic case).

\section{Ill-conditioned Jacobian}
We can switch to a conjugated gradient algorithm on $G$ since we have the knowledge of $\mygrad f$.
\end{document}

Nonetheless, we have a second order approximation of $f$
\begin{equation}
	f\left(\vec{X}+\vec{h}\right) \simeq f(\vec{X}) + \mytrn{\vec{F}} \mymat{J} \vec{h} +
	\frac{1}{2} \mytrn{\vec{h}} \mytrn{\mymat{J}}\mymat{J} \vec{h}
\end{equation}
which is as best as we are close to a root.
Since we assume that $\mymat{J}$ is ill-conditioned, we can use a singular value decomposition
of
\begin{equation}
	\mymat{J} = \mymat{U} \mymat{W} \mytrn{\mymat{V}}
\end{equation}
where $\mymat{W}$ is a diagonal matrix, $\mytrn{\mymat{U}}\mymat{U}=\mytrn{\mymat{V}}\mymat{V}=\mymat{I}_n$.
We obtain
\begin{equation}
	\mytrn{\mymat{J}}\mymat{J} = \mymat{V} \mymat{W}^2 \mytrn{\mymat{V}}.
\end{equation}
We transform $\mymat{W}$ into a non-singular matrix $\widetilde{\mymat{W}}$,
so that the minimum of $G$ is around the step
\begin{equation}
	\vec{h}_k = - \mymat{V} \left( \widetilde{\mymat{W}}^{-2} \mymat{W}\right) \mytrn{\mymat{U}} \vec{F}_k.
\end{equation}
We recognize the Newton's step when $\mymat{W}$ is well-conditioned.
The problem is to choose a good $\widetilde{\mymat{W}}$.
Especially, we should have
\begin{equation}
\begin{array}{rcl}
	\mygrad G_k \vec{h}_k & = &\mytrn{\vec{F}_k} \mymat{J} \vec{h_k} \\
	& = & - \mytrn{\vec{F}_k} \mymat{U} \left( \widetilde{\mymat{W}}^{-2} \mymat{W}^2 \right)\mytrn{\mymat{U}} \vec{F}_k < 0\\
\end{array}
\end{equation}
Since $\mygrad G_k$ is not zero, $\mytrn{\mymat{U}} \vec{F}_k$ is not zero.
Since $\left( \widetilde{\mymat{W}}^{-2} \mymat{W}^2 \right)$ is positive-definite,
\underline{$\vec{h}_k$ is a descent direction for $G$}.
A simple way to regularize is
\begin{equation}
	\widetilde{W}_i^2 = W_i^2 + \beta^2 
\end{equation}




\end{document}